4 Answers
4

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". The following extensive exposition by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this information is not complete even for systems as simple as quantum mechanics along a finite interval.

@David Bar Moshe Very interesting article, it is the first time I hear the terminology "isospectral potential". But I am also very concerned about the quality of the journal. Apeiron, anybody?!!
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RevoAug 13 '11 at 19:56

But in that case you get only half the spectrum of the usual harmonic oscillator, don't you?
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RevoAug 13 '11 at 20:10

@Revo is correct. E.g., the energy spectrum of the full harmonic oscillator is (1,3,5,7,...) all times $\hbar \omega_{full}/2$ and for the half harmonic oscillator is (3,7,11,15,...) times $\hbar \omega_{half}/2$. Just by changing $\omega_{half}$ by something independent of $n$, you can't get the terms to match up (e.g., for $n=0$; you'd have to let $3\omega_{full}=\omega_{half}$, for $n=1$ let $(7/3) \omega_{full}=\omega_{half}$, for $n=2$ let $(11/5)\omega_{full}=\omega_{half}$).
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dr jimbobAug 13 '11 at 20:51

@dr jimbob True, since the full harmonic oscillator solutions oscillate between even and odd, half the spectrum will not satisfy the boundary conditions at the origin of the harmonic oscillator potential with a hard wall at x=0, namely the wave function must vanish there.
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RevoAug 13 '11 at 21:09

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@Everybody: you can add a constant to a potential. It's still a potential. What is this discussion?
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Ron MaimonAug 14 '11 at 1:36

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@Marty--- the question is not physics, it is mathematics--- can you find Schrodinger operators with identical spectrum. This is a trivial example. More trivial examples are any potential and its translates and reflections, and a less trivial example is any potential and its supersymmetric conjugate, when the ground state doesn't break supersymmetry.
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Ron MaimonAug 14 '11 at 1:53

Example 1: If the potential $\Phi(x)=\Phi(-x)$ is an even function and is strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=2\Phi^{-1}(V)$ is twice the positive inverse branch of $\Phi$.

Example 2: If the potential has a hard wall $\Phi(x)=+\infty$ for $x<0$, and is strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=\Phi^{-1}(V)$ is the positive inverse branch of $\Phi$.

Example 3: If the potential $\Phi(x)$ is strongly monotonically decreasing for $x\leq0$ and strongly monotonically increasing for $x\geq 0$, then the accessible length $\ell(V)=\Phi_{+}^{-1}(V)-\Phi_{-}^{-1}(V)$ is the difference of the two inverse branch of $\Phi$.

In Example 1 and 2, if we would be able to determine the accessible length function $\ell(V)$, then we would also be able to generate the corresponding potential $\Phi(x)$ as OP asks.

This is repeating Jose's answer, again without explicitly stating that you are assuming that V(x)=V(-x) or a hard wall at negative x or some other way to remove the arbitrariness due to reflection symmetry.
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Ron MaimonOct 30 '11 at 1:23

No, I have divided the reconstruction(v1) of the potential $\Phi(x)$ into two parts: i) Firstly, calculating the accessible length function $\ell(V)$ from $N(E)$ via formula (3). It is remarkable that this is always possible, and this is the main claim. ii) Secondly, calculating the potential $\Phi(x)$ from $\ell(V)$, which is only possible under additional assumptions. The even case $\Phi(x)=\Phi(-x)$ is explicitly mentioned as an example.
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Qmechanic♦Oct 30 '11 at 11:51

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You are right, +1, I read it too quickly. It is an interesting answer, thanks.
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Ron MaimonOct 30 '11 at 20:27

another question.. once we have constructed the potential from $ N(E) $ can we prove that the Hamiltonian $ H=p^{2} + V(x) $ will satisfy the Gutzwiller Trace formula ? . I mean if Gutzwiller trace formula and the semiclassical reconstruction of this potential are seemingly related or not :)
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Jose Javier GarciaApr 9 '12 at 10:08

another comment, although $ N(E) $ is always postive, and if we conside even potential $ V(x)=V(-x) $, at least semiclassically is it possible to prove that $ V(x) $ will be positive if we know that $ l(V) $ is positive ??, thanks.
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Jose Javier GarciaOct 7 '12 at 8:18

you can search 'Wu and Sprung potential' Riemann Hypothesis as an inverse problem is treated there :)
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Jose Javier GarciaOct 9 '11 at 12:40

i believe that if the length function $ l(x) $ is increasing you can get the potential... for other cases you can always reflect the function 'l' trough the line $ y=x$ i used this spectral problem to find a suitable hamiltonian for the Riemann Hypothesis in the WKB case
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Jose Javier GarciaNov 10 '11 at 12:14